# Frequency of losing x times in a row per 100,000 hands

Discussion in 'General' started by D.G., May 1, 2007.

1. ### D.G.New Member

Hello. Where can I find out how many times I will lose 2, 3, 4, 5, 6, 7, 8, 9, 10,11,12,13,14,15 hands in a row per 10,000 hands, and per 100,000 hands?

I have scoured the web looking for this info, and I'm at a loss.

I would GREATLY appreciate any help.

2. ### KasiWell-Known Member

To get u started, if it were a coin toss, you'd lose x times in a row 0.5 to the x power.

Poke around wizardofodds.com. U'll find it.

3. ### SonnyWell-Known Member

You can also get the streak percentages for various games (number of decks, rules, etc.) here:

In the "Table" pulldown menu select "Streaks - Hands In A Row"

-Sonny-

4. ### D.G.New Member

Awesome, Sonny. That's EXACTLY what I was looking for. I only have one question, though. How do I find out how many total hands were played in this sim? That way, I will know how often you will get a streak of x per 10,000 hands, 100,000 hands, ect...

5. ### SonnyWell-Known Member

You can use the percentages for that. They will tell you the number of hands won/lost per hundred. Just multiply them by 100 to get the number of hands per 10,000 or by 1,000 to get the number of hands per 100,000.

-Sonny-

6. ### KasiWell-Known Member

Just wondering what u plan to do with the information? Design a betting system?

No need to reply if it's private.

7. ### D.G.New Member

Not private at all. And, yes, I'm trying to be the first person to ever design a progressive betting system that works, lol. I started out gambling hard core 4 years ago. I started counting cards with Brice Carlson's "Blackjack for Blood" counting system, but I didn't understand that counting doesn't change the # of hands you win/lose, so I was lucky to break even after 2 years of counting. The past 2 years I have gone broke playing NL Texas, and 7 card stud.

I feel like I have finally learned enough that I can make a run at a reliable income, and I feel like blackjack is the best game for that.

Here's my progressive betting strategy that I am trying to calculate-

Assume \$5 is your betting unit.

Lose 5 hands in a row, then Martingale it- \$25, \$50, \$100, ect....

My theory is that at some point, the frequency of losing x number of hands in a row will decrease to a point where when you finally do hit that big loser, you will have made more money up to that point than you lose.

To say it another way, I want to lose 5 hands in a row, then Martingale it up to a point where I will bust out on the 15th hand, with the theory that the money I have made on my previous winning sessions will be more than the money that I make on that one losing session, because the frequency of losing 15 hands in a row will be outwayed by all of the times I have succesfully Martingale'd.

My basis of this theory is that if you graph out the number of losing streaks that you will have over 100,000 hands ( x losing streaks of 3 hands, y losing streaks of 4 hands, z losing streaks of 5 hands, ect...), the graph is a curve, not a straight line.

I know that the chances of this working is slim, but I don't understand mathmatically why, so I am digging into the numbers.

8. ### D.G.New Member

Ok, so I think that I have it figured out. Out of every 100,000 hands played by this sim, there were-

LOSING STREAKS

10,544 streaks of 1
5,572 streaks of 2
2,954 streaks of 3
1,568 streaks of 4
832 streaks of 5
441 streaks of 6
234 streaks of 7
124 streaks of 8
66 streaks of 9
35 streaks of 10
19 streaks of 11
10 streaks of 12
5 streaks of 13
3 streaks of 14

Is that correct? It seems like there should be a lot more of the higher # losing streaks?

9. ### ihate17Well-Known Member

Just another Martingale but waiting till the sixth hand

Am I missing something here. If you win any of the 6th - 15th hands you will be even. If you lose the 15th hand you will be down \$25,600. Most times, of course, you will be even.
I know I sent you an PM in response to your sending me one, and stated that there have been several times where I have lost over 20 hands in a row and loses of 10 or more are just not that rare.
I also can not see where you will be picking up profits of more than \$5 at a time while subjecting yourself to a progression that might cause you to find a table that will accept a \$12,800 bet just to break even.

ihate17

10. ### D.G.New Member

Let's say you start the Martingale progression after the fifth loss in a row, so on the sixth bet you place a bet of \$25, then \$50, then \$100, then \$200, then \$400, and once you lose that \$400 bet, your session is over, \$800 down.

I think the average ratio of wins to losses is 45 wins per 100 hands, 49 losses, and 6 pushes. If you use this modified Martingale, the advantage is that you will have , on average, 3 streaks per hour of 6 or more losses in a row. If you cancel those losses out with a Martingale, you are effectively shifting the win/loss ratio into a state where you will win more hands than you lose.

The question that has to be answered is- Will shifting the win loss ratio in this manner make you more money than th amount of money that you will lose, once you do hit that tenth loss (or where ever you put the stop point).

So if you start the Martingale after 5 losses in a row, and you are prepared be wiped out (of that single session), after the 10th loss, will you have enough small streaks that shift the win loss ratio in your favor to outweigh the enevitable losing session?

11. ### EasyRhinoWell-Known Member

Well, a "regular martingale" trades an increased number of small wins for a smaller number of very large losses. There is no net change in house edge.

A "delayed martingale" just trades a smaller increase in the number of small wins, for very small number of very large loss. In fact, if you were playing a game where you never really won or lost more than 5 in a row (very possible in a short session), then your betting system wouldn't even take effect, and would be indistinguishable from flat-betting.

No net change in house edge.

Instead, you should look into sizing your blackjack bet in relation to the count, and your poker bet relative to the strength of your hand and the expectation of the opponent.

12. ### KasiWell-Known Member

And here I thought u just said in an earlier post that counting doesn't chnage the win/loss ratio lol.

This won't either.

What is your goal - win a certain number of \$5 units? Not lose? How will u make money before u lose 5 in a row? Using a Martingale to break even?

There are probably other systems that will win x a higher percentage of the time.

But good luck - just be able to afford to lose the money u commit to this with no regrets.

13. ### KnoxWell-Known Member

What about when you double down and lose? What if you lose two doubles in a row? I think that is the major weakness of the progression systems. If you say well I just won't double, you are lowering your EV significantly.

I had a discussion with a guy I work with last week who swears by his simple progression system. I think he is an imbecile and hopfully I won't have to work with him again anytime soon.

14. ### daciumWell-Known Member

I admire people like D.G who investigate things themselves. And it is the best way to learn, however in this case he should not be encourged as he is just wasting his time and ultimately time is life.

The sad fact here is that people like him just are not going to accept the fact that NO progression can win. This is mathematical FACT. It has been proven by mathematical formula thats cover EVERY possible progression. Many people do not understand how this can be so. But if you think about any progression is just a certain bet sized based on past outcomes, so a formula that covers every possible outcome and every possible bet size for any length of play from 0 to infinity does infact cover every possible progression that can occur, and the result is that they ALL fail.

The fact that he is looking for how often a streak occurs in 100,000 hands just show how his approach is completely wrong and extremely amature. If you can't calculate the probability of 14 losses in a row occurring in 100,000 hands, then you won't be able to understand the mathematical proofs that prove no possible progression can win, and this is sad point of this. How many people and how many years have we seen people come and go doing the same thing trying to figure out a progression 'that works'.

ps. lets assume its a coin toss game. chance of 14 losses in a row is 0.5^14 = 0.00006103515625, or 1 in 16384. You are playing 100,000-13= 99987 times. So you expect it to occur 99987/16384 = 6.1 times.

I suspect the reason your sim seemed to show less losses is simply because of two reasons, either you are incorrectly counting a push as the end of a loosing streak (a loosing streak does not end until a win occurs). Or because quite simply we have not calculated the variance. The variance here is quite likely extremely high. To get a real accurate number you would have to run the 100,000 hand simulation 100,000 times...

15. ### KasiWell-Known Member

I hope u don't mind me posting this quote from the General forum but I figure it might as well end up here.

Just to satisfy my own curiosity could u sim say a 6D AC game, u pick the rules.
Assume I always have enuf bankroll to play 100 hands. If I play 100 hands without ever being 1 unit up, I quit. If I get up 1 unit or more, I quit and start a new session. I flat bet every hand with BS. Re-shuffle after every hand.

Run as many sessions as u feel like - maybe 100,000 or so?. Just curious as to the number of winning sessions, amongst other things.

If u want to change the maximum of 100 hands per session to 500 or 1000 feel free to do so.

Thanks. And no problem if this is a big pain in the u-know-what lol.

As far as the losing streaks that DG posted go, I don't see anything wrong with the number of losses - what don't u like about it?

16. ### taipafanMember

Just a guess...a loosing streak starts only after a win, only half of 99987 make the chances.

17. ### aslanWell-Known Member

Modification, smodification...!

I tried several modifications thinking there might be some way to beat the odds. It ain't gonna happen! But when you add the house edge on top of the fact that NO MARTINGALE, modified or not, can ever be a winner---well, the results are devastating.

I first looked at the math. You're right--it just isn't viable. But then I thought, as so many do, that maybe the math was missing something. So I devised a spreadsheet that would run 10,000 hands at a time, trying all sorts of modifications. You simply cannot outrun the odds. This hands-on experiment showed you simply book loser after loser in the long haul. It is a hopeless pursuit, a pipe dream.

18. ### KasiWell-Known Member

Try flat-betting \$1/hand for 10000 hands with a \$1000 bankroll.

Maybe u lose \$43.

Now flat-bet a \$43 unit until u are at least one unit ahead and your profit is now zero (or a little more like 1.5 units or 2 units if u happened to wi a BJ or double, etc.)

Then play another 10000 hands. And that's an aggressive scheme for for beating the HA.

How many hands did u last?

Even if u flat-bet \$1, you'd probably still be ahead in actual dollars (let alone only exceeding the HA) after 10000 hands almost 2/3 of the time (depending on the rules, etc.)

In other words, while of course u can never beat the house advantage in the long run, you might have a real good chance over your lifetime.

Of course, u won't be rich lol.

19. ### daciumWell-Known Member

What do you even mean? You still are not making much sense to figure out what you are trying to say.

Do you mean you flat bet \$1/hand for 10,000 hands.

Then you bet whatever you lost until you are ahead some amount of money.

Then you start over.

This will not win you money. When you increase your bet you will just loose more money, you still need luck to get back \$43 you are down. For example 50.5% of the time you loose the first bet and then end up sinking lower and lower.

20. ### aslanWell-Known Member

Not a bad idea. Of course, that's not a martingale. And right! you won't exactly get rich!